Specific Solution for a Differential Equation

• Oct 20th 2008, 08:35 PM
veronicak5678
Specific Solution for a Differential Equation
Find the solution of the diff eq that satisfies the given initial condition.

y'tan x = a + y, y(pi/4) = a, 0 < x < pi/2

y' = (a+y)/tanx

dy/ (a+y) = cot x dx

Integrate...

ln (a + y ) = ln (sinx ) + c

I think to make this work, c will be 2a- root2/2, so my answer is
looks like

y = sinx - a + 2a - root2 /2

Is this correct?
• Oct 20th 2008, 08:42 PM
Chris L T521
Quote:

Originally Posted by veronicak5678
Find the solution of the diff eq that satisfies the given initial condition.

y'tan x = a + y, y(pi/4) = a, 0 < x < pi/2

y' = (a+y)/tanx

dy/ (a+y) = cot x dx

Integrate...

ln (a + y ) = ln (sinx ) + c

I think to make this work, c will be 2a- root2/2, so my answer is
looks like

y = sinx - a + 2a - root2 /2

Is this correct?

When solving for y, you should end up with $a+y=e^{\ln(\sin x)+C}\implies y=C\sin x-a$

When you apply the initial condition, you should get $C=2\sqrt{2}a$

Thus, $\color{red}\boxed{y=2\sqrt{2}a\sin x-a}$

Does this make sense?

--Chris
• Oct 20th 2008, 08:54 PM
veronicak5678
No... could you show me how to solve for y?
• Oct 20th 2008, 08:59 PM
Chris L T521
Quote:

Originally Posted by veronicak5678
No... could you show me how to solve for y?

Your work is good up to here:

$\ln(a+y)=\ln(\sin x)+C$

To get rid of the logarithm on the left side of the equation, apply the exponential function:

$\ln(a+y)=\ln(\sin x)+C\implies e^{\ln(a+y)}=e^{\ln(\sin x)+C}\implies a+y=e^{\ln(\sin x)+C}$

Thus,

$a+y=e^{\ln(\sin x)+C}\implies a+y=e^{\ln(\sin x)}e^C\implies a+y=Ce^{\ln(\sin x)}$; $e^C=C$

Finally,

$a+y=Ce^{\ln(\sin x)}\implies a+y=C\sin x\implies y=C\sin x-a$

Now plug in the initial condition to find C.

Does this clarify things?

--Chris
• Oct 20th 2008, 09:08 PM
veronicak5678
Got it. I would never have known to do that! Thanks a lot.